1. 69.8 m
The vertical motion of the shell is a uniformly accelerated motion, with constant acceleration [tex]g=-9.81 m/s^2[/tex] towards the ground (acceleration due to gravity).
At the point of maximum height, the velocity of the projectile is zero:
v = 0
So we can find the maximum height by using the equation:
[tex]v^2 -u^2 = 2gd[/tex]
where
u = 37 m/s is the initial velocity
d is the maximum heigth
Solving for d,
[tex]d=\frac{v^2-u^2}{2g}=\frac{0^2-(37 m/s)^2}{2(-9.81 m/s^2)}=69.8 m[/tex]
2)
A projectile motion consists of two separate motions:
- A uniform motion along the x-direction, with constant velocity given by
[tex]v_x = v_0 cos \theta[/tex]
where [tex]v_0[/tex] is the initial velocity, and [tex]\theta[/tex] the angle of launch
- A unformly accelerated motion along the y-direction, with initial velocity
[tex]v_y = v_0 sin \theta[/tex]
and constant acceleration
[tex]g=-9.81 m/s^2[/tex] (acceleration due to gravity) towards the ground.
The horizontal position of the projectile at time t is given by
[tex]x(t) = v_x t[/tex]
while the vertical position is given by
[tex]y(t) = y_0 + v_0 sin \theta t + \frac{1}{2}gt^2[/tex]
where [tex]y_0[/tex] is the initial height of the projectile.
